Respuesta :
Since the x-coordinates are equal, the ellipse is vertical. Thus, the standard equation of the ellipse is as follows:
[tex]\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1[/tex]Since the x-coordinates of the foci and the vertex are the same, the value of h must be 0.
Equate the y-coordinates of the foci to the following.
[tex]\begin{gathered} k+c=4 \\ k-c=-4 \end{gathered}[/tex]Add the two equations and then solve for k.
[tex]\begin{gathered} 2k=0 \\ k=0 \end{gathered}[/tex]Thus, the center is at (0,0) or the origin.
To solve for c, substitute the value of k into k+c=4.
[tex]\begin{gathered} k+c=4 \\ 0+c=4 \\ c=4 \end{gathered}[/tex]To solve for a, equate the y-coordinate of the vertex to k+a.
[tex]\begin{gathered} k+a=6 \\ 0+a=6 \\ a=6 \end{gathered}[/tex]Solve for a².
[tex]a^2=6^2=36[/tex]To solve for b², substitute the value of a and c into the following equation.
[tex]b^2=a^2-c^2[/tex]Thus, we obtain the following:
[tex]\begin{gathered} b^2=6^2-4^2 \\ b^2=36-16 \\ b^2=20 \end{gathered}[/tex]Substitute the values into the equation of the ellipse. Thus, we obtain the following:
[tex]\begin{gathered} \frac{(x-0)^2}{20}+\frac{(y-0)^2}{36^{}}=1 \\ \frac{x^2}{20}+\frac{y^2}{36^{}}=1 \end{gathered}[/tex]Otras preguntas
